Answer:
Jim's error is " He did not multiply Three-fifths by 2 before applying the power "
Step-by-step explanation:
Jim's evaluating expression is 
To verify Jim's error :
Jim's steps are
Therefore 
Jim's error is " He did not multiply Three-fifths by 2 before applying the power "
That is the corrected steps are
( using the property 
)
Answer:
C
Step-by-step explanation:
I might be wrong. I hope it helps:)
The answer is C, 75. If you substitute in 75 you will get 32
Answer:
π/6 [37^(³/₂) − 1] ≈ 117.3187
Step-by-step explanation:
The surface area is:
S = ∫ 2π (x − 0) √(1 + (dx/dy)²) dy
0 ≤ x ≤ 3, so -4 ≤ y ≤ 5.
Find dx/dy.
y = 5 − x²
x² = 5 − y
x = √(5 − y)
dx/dy = ½ (5 − y)^(-½) (-1)
dx/dy = -½ (5 − y)^(-½)
(dx/dy)² = ¼ (5 − y)^(-1)
(dx/dy)² = 1 / (4 (5 − y))
Plug in:
S = ∫₋₄⁵ 2π x √(1 + 1 / (20 − 4y)) dy
S = ∫₋₄⁵ 2π √(5 − y) √(1 + 1 / (4 (5 − y))) dy
S = ∫₋₄⁵ 2π √((5 − y) + 1/4)) dy
S = ∫₋₄⁵ 2π √(5.25 − y) dy
If u = 5.25 − y, then du = -dy.
S = ∫ 2π √u (-du)
S = -2π ∫ √u du
S = -2π (⅔ u^(³/₂))
S = -4π/3 u^(³/₂)
Substitute back:
S = -4π/3 (5.25 − y)^(³/₂)
Evaluate between y=-4 and y=5.
S = [-4π/3 (5.25 − 5)^(³/₂)] − [-4π/3 (5.25 − -4)^(³/₂)]
S = -4π/3 (0.25)^(³/₂) + 4π/3 (9.25)^(³/₂)
S = π/6 [37^(³/₂) − 1]
S ≈ 117.3187
Answer:
The probability that the two rats are from the first litter is 14.28%, and the probability that the two rats are from the second litter is 34.28%.
Step-by-step explanation:
Since a cage holds two litter of rats, and one litter comprises one female and five males, while the other litter comprises seven females and two males, and a random selection of two rats is done, to find the probability that the two rats are from the same litter the following calculation must be performed:
6/15 x 5/14 = 0.1428
9/15 x 8/14 = 0.3428
Therefore, the probability that the two rats are from the first litter is 14.28%, and the probability that the two rats are from the second litter is 34.28%.
